For a constant perturbation of the form $$\hat{H'}(t)=\hat{V}\theta(t)$$ to a time-independent Hamiltonian $\hat{H}_0$, the transition probability at time $t$ from an eigenstate $|i\rangle$ of $\hat{H}_0$ with energy $E_i$ at $t=0$ to a final state $|f\rangle$ ($\neq |i\rangle$) is given by $$P_{i\to f}(t;\omega_{fi})=|V_{fi}|^2\left[\frac{\sin(\omega_{fi}t/2)}{\omega_{fi}/2}\right]^2$$ where $\omega_{fi}=(E_f-E_i)/\hbar$. At a fixed $t$, the transition probability $P_{i\to f}(t;\omega_{fi})$ peaks at $\omega_{fi}=0$ i.e., for $E_f=E_i$. Why does a constant perturbation favour the transition at $\omega_{fi}=0$? Is it because the constant perturbation has only $\omega=0$ frequency component? But then why don't we see a delta function peak at $\omega=0$?

enter image description here

  • $\begingroup$ If perturbation is constant, it has zero matrix elements between any pait of eigenstates of the unperturbed Hamiltonian, hence no transitions. $\endgroup$
    – Roger V.
    Commented Jun 29, 2021 at 15:43
  • $\begingroup$ Related: physics.stackexchange.com/q/89402/2451 $\endgroup$
    – Qmechanic
    Commented Jun 29, 2021 at 15:44
  • $\begingroup$ @RogerVadim This is not true. There is no reason why $V$ should be diagonal in the eigenbasis of $H_0$. This is from Sakurai. $\endgroup$ Commented Jun 29, 2021 at 16:02
  • $\begingroup$ @mithusengupta123 perhaps, you should clarify what constant means in your question: it is not constant in time (it is a sudden perturbation that turns on at $t=0$), and, according to your comment, it is not a constant. $\endgroup$
    – Roger V.
    Commented Jun 29, 2021 at 16:05
  • 1
    $\begingroup$ Chapter 5.6, Page 328 of Modern Quantum Mechanics Revised Edition, J.J. Sakurai, for those interested. $\endgroup$ Commented Jun 29, 2021 at 16:15

2 Answers 2


Your perturbation is by no means constant as it has a step function, which has all frequencies in its Fourier transformation. For such violent step-function perturbations you should use the "sudden" approximation in which the transition amplitude is simply the overlap between the eigenstates of $H_0$ and $H_0+V$.

  • $\begingroup$ This is in Sakurai, under a section titled constant perturbation. Your answer does not explain why the peak occurs at $\omega_{fi}=0$. $\endgroup$ Commented Jun 29, 2021 at 16:42

Note that your plot assumes that $V_{if}$ is non-zero. This is telling you that if the perturbation connects degenerate states (such that $\omega_{if} = E_f-E_i$ is zero and $V_{if}\neq 0$), then the transition between these states is favoured over the others.

The transition to other states would cost more energy (than zero) so it is less likely.

If the perturbation is diagonal in the degenerate subspace, or if there is no degeneracy, then there are no $\omega_{if}$ and $V_{if}$ that satisfy the conditions above simultaneously.

  • $\begingroup$ I didn't get it. In general, the given perturbation will connect both degenerate as well as nondegenerate states. What is so special about the perturbation that it favors transition between degenerate states? Not clear. $\endgroup$ Commented Jun 29, 2021 at 16:54
  • $\begingroup$ It is not the perturbation per se that favors the degenerate states, it is conservation of energy. If the initial system is in a given energy eigenstate, it will need to absorb or give up energy in order to jump to states that are non-degenerate to it. This breaks conservation of energy, but that is ok because we are doing everything at finite times and the uncertainty relations warned us of this. As a consequence, though, the higher the energy jump the bigger the "breaking", so the less likely it should happen. $\endgroup$ Commented Jun 29, 2021 at 17:12
  • $\begingroup$ Of course, this is a general behavior and the details of the perturbation can change this. That is why the matrix elements $V_{if}$ modulate the probability amplitude and if they are zero, that transition does not happen. $\endgroup$ Commented Jun 29, 2021 at 17:14
  • $\begingroup$ "This breaks conservation of energy" I'm not sure if this is true. The perturbation is there to supply any energy. If the perturbation cannot supply adequate energy, there must be a reason. Moreover, although before the perturbation the system was in an eigenstate of $H_0$, after the perturbation acts, the system is not in an energy eigenstate. So statements about energy conservation are a bit shaky (because energy is not well-defined), in my opinion. $\endgroup$ Commented Jun 29, 2021 at 17:29
  • $\begingroup$ Fair point. I will leave this answer here as is until I think of a counter-argument or someone comes up with a better explanation. $\endgroup$ Commented Jun 29, 2021 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.